Solve the inequality. (Enter your answer using interval notation.) \[ x^{3}+6 x^{2}-36 x \leq 216 \] Graph the solution set.

To solve the inequality \( x^{3}+6 x^{2}-36 x \leq 216 \), we first rearrange it to standard form: \[ x^{3}+6 x^{2}-36 x - 216 \leq 0. \] Next, let's find the roots of the corresponding equation \( x^{3}+6 x^{2}-36 x - 216 = 0 \). Using synthetic division or trial and error, we can find one root; let's say \( x = 6 \) is a root. Now, we can factor the polynomial: 1. Divide \( x^{3}+6x^{2}-36x-216 \) by \( x-6 \): - You should find \( x^{3} + 6x^{2} - 36x - 216 = (x - 6)(x^{2} + 12x + 36) \). 2. The quadratic can be rewritten as: - \( x^{2} + 12x + 36 = (x + 6)^{2} \). Thus, we can express the entire polynomial as: \[ (x - 6)(x + 6)^{2} \leq 0. \] Now we determine the roots: - The roots are \( x = 6 \) (single root) and \( x = -6 \) (double root). Next, we identify the intervals to test based on these roots: 1. \( (-\infty, -6) \) 2. \( (-6, 6) \) 3. \( (6, \infty) \) **Testing the intervals**: - For \( x < -6 \) (say \( x = -7 \)): - \( (-7 - 6)((-7 + 6)^{2}) \) is positive. - For \( -6 < x < 6 \) (say \( x = 0 \)): - \( (0 - 6)((0 + 6)^{2}) \) is negative. - For \( x > 6 \) (say \( x = 7 \)): - \( (7 - 6)((7 + 6)^{2}) \) is positive. **Conclusion**: The inequality is satisfied where it is less than or equal to zero, which occurs in the interval \([-6, 6]\). Putting this all together, the solution in interval notation is: \[ \boxed{[-6, 6]} \] To graph the solution, you would plot a closed circle (indicating that -6 and 6 are included in the solution set) along the x-axis, marking that everything between these points satisfies the inequality.